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17 tháng 10 2018

b: =>(x-5)2+6(x-5)=0

=>(x-5)(x+1)=0

=>x=5 hoặc x=-1

c: \(\Leftrightarrow x^2-x+2=2\)

=>x(x-1)=0

=>x=0(loại) hoặc x=1(nhận)

a) Ta có: \(\left|x^2-x+2\right|-3x-7=0\)

\(\Leftrightarrow\left|x^2-x+2\right|=3x+7\)

\(\Leftrightarrow x^2-x+2=3x+7\)(Vì \(x^2-x+2>0\forall x\))

\(\Leftrightarrow x^2-x+2-3x-7=0\)

\(\Leftrightarrow x^2-4x-5=0\)

\(\Leftrightarrow x^2-5x+x-5=0\)

\(\Leftrightarrow x\left(x-5\right)+\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

Vậy: S={5;-1}

6 tháng 3 2021

bạn giải giúp mình câu b nữa với

mai mình phải nộp bài rồi!!!khocroi

\(a,\\ \Leftrightarrow3x-x=10-6\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\\ b,\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

22 tháng 3 2022

\(a,\Leftrightarrow3x-x=10-6\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=\dfrac{4}{2}=2\)

Vậy phương trình có tập nghiệm S = \(\left\{2\right\}\)

\(b,\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow x+1=0\)           hoặc            \(\Leftrightarrow x-2=0\)       

\(\Leftrightarrow x=-1\)                                                     \(\Leftrightarrow x=2\)

Vậy phương trình có tập nghiệm S = \(\left\{-1;2\right\}\)

13 tháng 9 2021

\(a,\Leftrightarrow x^2+2x+1+2x+3-2\sqrt{2x+3}+1=0\\ \Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1\\2x+3=1\end{matrix}\right.\Leftrightarrow x=-1\left(N\right)\)

13 tháng 9 2021

\(b,\Leftrightarrow3x^2+3x-2\sqrt{x^2+x}=0\left(x\le-1;x\ge0\right)\\ \Leftrightarrow3x\left(x-1\right)-2\sqrt{x\left(x+1\right)}=0\\ \Leftrightarrow\sqrt{x\left(x+1\right)}\left(3\sqrt{x\left(x-1\right)}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x\left(x-1\right)=0\\\sqrt{x\left(x-1\right)}=\dfrac{2}{3}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x^2-x-\dfrac{4}{9}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\9x^2-9x-4=0\left(1\right)\end{matrix}\right.\)

\(\Delta\left(1\right)=81-4\left(-4\right)\cdot9=225\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{9-15}{18}\\x=\dfrac{9+15}{18}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(N\right)\\x=1\left(N\right)\\x=-\dfrac{1}{3}\left(L\right)\\x=\dfrac{4}{3}\left(N\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{4}{3}\end{matrix}\right.\)

16 tháng 1 2021

a) \(x^2+2x=\left(x-2\right).3x\)

\(\Leftrightarrow x^2+2x=3x^2-6x\)

\(\Leftrightarrow x^2+2x-3x^2+6x=0\)

\(\Leftrightarrow-2x^2+8x=0\)

\(\Leftrightarrow-2x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy S = {0;4}

b) \(x^3+x^2-x-1=0\)

\(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^2-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\mp1\end{matrix}\right.\)

Vậy: S = {-1; 1}

c) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)

\(\Leftrightarrow\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]=40\)

\(\Leftrightarrow\left(x^2+5x+x+5\right)\left(x^2+4x+2x+8\right)=40\)

\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)

Đặt x2 + 6x + 5 = t

\(\Leftrightarrow t.\left(t+3\right)=40\)

\(\Leftrightarrow t^2+3t=40\)

\(\Leftrightarrow t^2+2.t.\dfrac{3}{2}+\dfrac{9}{4}=\dfrac{169}{4}\)

\(\Leftrightarrow\left(t+\dfrac{3}{2}\right)^2=\dfrac{169}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}t+\dfrac{3}{2}=\dfrac{13}{2}\\t+\dfrac{3}{2}=-\dfrac{13}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{13}{2}-\dfrac{3}{2}=\dfrac{10}{2}=5\\t=-\dfrac{13}{2}-\dfrac{3}{2}=-\dfrac{16}{2}=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x+5=5\\x^2+6x+5=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x=0\\x^2+6x+13=0\end{matrix}\right.\)

Mà: \(x^2+6x+13=x^2+2.x.3+9+4=\left(x+3\right)^2+4\ne0\)

=> x2 + 6x = 0

<=> x. (x + 6) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

Vậy S = {0; -6}

 

 

a) Ta có: \(x^2+2x=\left(x-2\right)\cdot3x\)

\(\Leftrightarrow x\left(x+2\right)-3x\left(x-2\right)=0\)

\(\Leftrightarrow x\left[\left(x+2\right)-3\left(x-2\right)\right]=0\)

\(\Leftrightarrow x\left(x+2-3x+6\right)=0\)

\(\Leftrightarrow x\left(-2x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-2x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-2x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy: S={0;4}

b) Ta có: \(x^3+x^2-x-1=0\)

\(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\cdot\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\cdot\left(x-1\right)\cdot\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

Vậy: S={-1;1}

c) Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)

\(\Leftrightarrow\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)-40=0\)

\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)-40=0\)

\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)+40-40=0\)

\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)=0\)

\(\Leftrightarrow\left(x^2+6x\right)\left(x^2+6x+13\right)=0\)

\(\Leftrightarrow x\left(x+6\right)\left(x^2+6x+13\right)=0\)

mà \(x^2+6x+13>0\forall x\)

nên \(x\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

Vậy: S={0;-6}